state space is non-empty

In this entry we prove the existence of states for every $C^{*}$-algebra (http://planetmath.org/CALgebra).

Theorem - Let $𝒜$ be a $C^{*}$-algebra. For every self-adjoint (http://planetmath.org/InvolutaryRing) element $a\in 𝒜$ there exists a state $\psi$ on $𝒜$ such that $|\psi (a)|=\parallel a\parallel$.

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Proof : We first consider the case where $𝒜$ is unital (http://planetmath.org/Ring), with identity element $e$.

Let $ℬ$ be the $C^{*}$-subalgebra generated by $a$ and $e$. Since $a$ is self-adjoint, $ℬ$ is a comutative $C^{*}$-algebra with identity element.

Thus, by the Gelfand-Naimark theorem, $ℬ$ is isomorphic to $C(X)$, the space of continuous functions $X⟶ℂ$ for some compact set $X$.

Regarding $a$ as an element of $C(X)$, $a$ attains a maximum at a point $x_0\in X$, since $X$ is compact. Hence, $\parallel a\parallel =|a(x_0)|$.

The evaluation function at $x_0$,

$e{v}_{x_0}:C(X)⟶ℂ$

$e{v}_{x_0}(f):=f(x_0)$

is a multiplicative linear functional of $C(X)$. Hence, $\parallel e{v}_{x_0}\parallel =1$ and also $|e{v}_{x_0}(a)|=|a(x_0)|=\parallel a\parallel$.

We can now extend $e{v}_{x_0}$ to a linear functional $\psi$ on $𝒜$ such that $\parallel \psi \parallel =\parallel e{v}_{x_0}\parallel =1$, using the Hahn-Banach theorem.

Also, $\psi (e)=e{v}_{x_0}(e)=1$ and so $\psi$ is a norm one positive linear functional, i.e. $\psi$ is a state on $𝒜$.

Of course, $\psi$ is such that $|\psi (a)|=|e{v}_{x_0}(a)|=\parallel a\parallel$.

In case $𝒜$ does not have an identity element we can consider its minimal unitization $\widetilde{𝒜}$. By the preceding there is a state $\tilde{\psi }$ on $\widetilde{𝒜}$ satisfying the required . Now, we just need to take the restriction (http://planetmath.org/RestrictionOfAFunction) of $\tilde{\psi }$ to $𝒜$ and this restriction is a state in $𝒜$ satisfying the required . $\mathrm{\square }$